Quadratic functions are a turning point in Algebra 2. Students move from solving simple equations to interpreting graphs, comparing forms, analyzing maximum and minimum values, and choosing between factoring, completing the square, and the quadratic formula. A strong review should mix all of those skills instead of practicing one method at a time.
This review edition includes worked examples, original practice problems, and a compact answer key. Use it before a unit test, final exam, college placement review, or a return to Algebra 2 after a break.
Quick Reference: Three Forms of a Quadratic
| Form | What it shows quickly | Example |
|---|---|---|
| Standard form | y-intercept and formula setup | f(x) = ax^2 + bx + c |
| Vertex form | vertex and direction | f(x) = a(x – h)^2 + k |
| Factored form | zeros or x-intercepts | f(x) = a(x – r1)(x – r2) |
In standard form, the x-coordinate of the vertex is x = -b/(2a). In vertex form, the vertex is (h, k). In factored form, the zeros are r1 and r2.
Worked Example 1: Find the Vertex
Problem: Find the vertex and axis of symmetry for f(x) = x^2 – 10x + 17.
Solution: Here a = 1 and b = -10. The vertex x-coordinate is x = -b/(2a) = -(-10)/(2 x 1) = 5. Then f(5) = 25 – 50 + 17 = -8. The vertex is (5, -8) and the axis of symmetry is x = 5.
Worked Example 2: Convert to Vertex Form
Problem: Write f(x) = x^2 + 6x + 4 in vertex form.
Solution: Complete the square: x^2 + 6x + 4 = (x^2 + 6x + 9) – 5 = (x + 3)^2 – 5. So the vertex form is f(x) = (x + 3)^2 – 5.
Worked Example 3: Choose a Solving Method
Problem: Solve 2x^2 – 7x – 4 = 0.
Solution: This factors: 2x^2 – 7x – 4 = (2x + 1)(x – 4). Set each factor equal to zero. Then 2x + 1 = 0 gives x = -1/2, and x – 4 = 0 gives x = 4. The solutions are x = -1/2 and x = 4.
Practice Set A: Graphs and Forms
- Find the vertex of f(x) = x^2 – 8x + 11.
- Find the axis of symmetry for g(x) = -2x^2 + 12x – 7.
- Write h(x) = x^2 – 4x + 9 in vertex form.
- A parabola has vertex (2, -3) and passes through (4, 5). Write its equation in vertex form.
- For f(x) = -3(x + 1)^2 + 6, state the vertex and whether the graph opens up or down.
Practice Set B: Zeros and Solutions
- Solve x^2 – 9x + 20 = 0 by factoring.
- Solve 3x^2 + 5x – 2 = 0 by factoring.
- Use the quadratic formula to solve x^2 + 4x – 1 = 0.
- How many real solutions does 2x^2 – 4x + 7 = 0 have?
- Solve x^2 – 6x + 1 = 0 by completing the square.
Practice Set C: Applications
- A ball’s height is h(t) = -16t^2 + 48t + 5. At what time does it reach its maximum height?
- The area of a rectangle is modeled by A(x) = -x^2 + 18x. What value of x gives the maximum area?
- The profit from selling x items is P(x) = -2x^2 + 80x – 300. How many items maximize profit?
- A quadratic has zeros 3 and 11 and opens upward. Write one possible equation.
- A parabola touches the x-axis at x = -4. What does that tell you about its discriminant?
Answer Key
1. x = 4, f(4) = -5, so vertex (4, -5). 2. x = -b/(2a) = -12/(2 x -2) = 3. 3. (x – 2)^2 + 5. 4. y = a(x – 2)^2 – 3. Use (4, 5): 5 = 4a – 3, so a = 2. Equation: y = 2(x – 2)^2 – 3. 5. Vertex (-1, 6); opens down.
6. (x – 4)(x – 5) = 0, so x = 4 or 5. 7. (3x – 1)(x + 2) = 0, so x = 1/3 or -2. 8. x = [-4 +/- sqrt(16 + 4)]/2 = -2 +/- sqrt(5). 9. Discriminant = 16 – 56 = -40, so no real solutions. 10. (x – 3)^2 = 8, so x = 3 +/- 2sqrt(2).
11. t = -48/[2(-16)] = 1.5 seconds. 12. x = -18/[2(-1)] = 9. 13. x = -80/[2(-2)] = 20 items. 14. y = (x – 3)(x – 11), or any positive multiple. 15. The discriminant is 0 because there is exactly one real root.
Common Mistakes to Fix
- Reading vertex form backward: In (x – h)^2 + k, the vertex uses h, not -h.
- Using the quadratic formula too late: If factoring is not obvious after 20 seconds, switch methods.
- Forgetting the negative leading coefficient: If a is negative, the vertex is a maximum, not a minimum.
- Dropping the +/- sign: Any square-root solving step can create two solutions.
How to Turn This Into a Two-Day Review
On Day 1, focus only on forms and graphs. Rewrite three quadratics from standard form to vertex form, identify the vertex, and sketch the direction of opening. Then take three factored-form examples and identify the zeros without expanding. The purpose is to see what each form reveals quickly.
On Day 2, focus on solving. Mix factoring, completing the square, and the quadratic formula in one set. Before solving each equation, write the method you plan to use. If factoring is fast and clean, use it. If not, switch to completing the square or the quadratic formula. This choice-making skill is what many Algebra 2 tests are really measuring.
Next Study Step
If these problems feel uneven, do not retake a full test immediately. Make three short review sets: one for vertex and graph interpretation, one for solving equations, and one for applications. ViewMath Algebra 2 and college-placement resources can help students move from guided examples into mixed practice with full explanations.
